C++ 计算保存在文本文件中的 2 个纬度和经度之间的距离？
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Calculating the distance between 2 latitudes and longitudes that are saved in a text file?
提问by Ange
I've looked around and still couldn't find anything to help me really! I've written a program to calculate the distance between 2 cities using their Latitudes and Longitudes, the cities details are saved in a file then loaded in my program into a BST! So far everything works fine except when i run the codes that is suppose to calculate the distance i get the same answer for every cities! Am not quite sure as to why i keep getting the same answer for every cities! Please help point me into the right direction?
我环顾四周，仍然找不到任何可以真正帮助我的东西！我已经编写了一个程序来使用它们的纬度和经度来计算两个城市之间的距离，城市的详细信息保存在一个文件中，然后在我的程序中加载到 BST 中！到目前为止，一切正常，除了当我运行假设计算距离的代码时，我对每个城市都得到了相同的答案！我不太确定为什么我对每个城市都得到相同的答案！请帮我指出正确的方向？
here are the codes to calculate the distance
这是计算距离的代码
#include <cmath>
#define pi 3.14159265358979323846
string userResponse;
float globalLat1, globalLon1, globalLat2, globalLon2;
for(int j= 0; j < 2; j++){
string whatever;
if (j==0){
bool hasbeenfound = false;
do{
//ask the user to enter their first city of their choice
whatever = "first ";
cout << "Enter your " + whatever + "City" << endl;
cout << "-------------------" << endl;
cin >> userResponse;
cout << endl;
if (Cities->search(userResponse)) //check if the entered city already exist
{
hasbeenfound = true;
}
else{
cout << "City not Found" << endl;
cout << endl;
}
//globalCity1 = Cities->sRootName;
globalLat1 = Cities->sLatitude;
globalLon1 = Cities->sLongitude;
}
while(hasbeenfound == false); //while the entered city hasn't been found, repeat the process
}else
{
bool hasbeenfound = false;
do{
//ask the user to enter their second city of their choice
whatever = "second ";
cout << endl;
cout << "Enter your " + whatever + "City" << endl;
cout << "-------------------" << endl;
cin >> userResponse;
cout << endl;
if (Cities->search(userResponse)) //check if the entered city already exist
{
hasbeenfound = true;
}
else{
cout << "City not Found" << endl;
}
//globalCity2 = Cities->sRootName;
globalLat2 = Cities->sLatitude;
globalLon2 = Cities->sLongitude;
}
while(hasbeenfound == false); //while the entered city hasn't been found, repeat the process
}
}
// This function converts decimal degrees to radians
double deg2rad(double deg) {
return (deg * pi / 180);
};
// This function converts radians to decimal degrees
double rad2deg(double rad) {
return (rad * 180 / pi);
};
//distance calculations
cout << endl;
distan = sin(globalLat1)) * sin(deg2rad(globalLat2)) + cos(deg2rad(globalLat1)) * cos(deg2rad(globalLat2)) * cos(globalLon2 - globalLon1);
distan = rad2deg(distan);
distan = distan * 60 * 1.1515;
distan = (6371 * pi * distan)/180;
cout << "The Distance between the to cities is: " << distan << " kilometers" << endl;
回答by MrTJ
As it is said, Haversine formula is your answer:
如前所述，Haversine 公式就是您的答案：
#include <math.h>
#include <cmath>
#define earthRadiusKm 6371.0
// This function converts decimal degrees to radians
double deg2rad(double deg) {
return (deg * M_PI / 180);
}
// This function converts radians to decimal degrees
double rad2deg(double rad) {
return (rad * 180 / M_PI);
}
/**
* Returns the distance between two points on the Earth.
* Direct translation from http://en.wikipedia.org/wiki/Haversine_formula
* @param lat1d Latitude of the first point in degrees
* @param lon1d Longitude of the first point in degrees
* @param lat2d Latitude of the second point in degrees
* @param lon2d Longitude of the second point in degrees
* @return The distance between the two points in kilometers
*/
double distanceEarth(double lat1d, double lon1d, double lat2d, double lon2d) {
double lat1r, lon1r, lat2r, lon2r, u, v;
lat1r = deg2rad(lat1d);
lon1r = deg2rad(lon1d);
lat2r = deg2rad(lat2d);
lon2r = deg2rad(lon2d);
u = sin((lat2r - lat1r)/2);
v = sin((lon2r - lon1r)/2);
return 2.0 * earthRadiusKm * asin(sqrt(u * u + cos(lat1r) * cos(lat2r) * v * v));
}
回答by Neel Basu
Using boost.geometry
使用 boost.geometry
typedef boost::geometry::model::point<
double, 2, boost::geometry::cs::spherical_equatorial<boost::geometry::degree>
> spherical_point;
spherical_point p(lon1_degree, lat1_degree);
spherical_point q(lon2_degree, lat2_degree);
double dist = boost::geometry::distance(p, q);
double const earth_radius = 6371.0; // Km
double dist_km = dist*earth_radius;
回答by ingconti
for people who need in swift:
对于需要快速的人：
// Haversine formula:
func deg2rad(_ deg: Double) ->Double {
return deg * Double.pi / 180.0
}
func distanceEarth(lat1d: Double, lon1d: Double, lat2d: Double, lon2d: Double) ->Double {
let earthRadiusKm = 6371.0
let lat1r = deg2rad(lat1d);
let lon1r = deg2rad(lon1d);
let lat2r = deg2rad(lat2d);
let lon2r = deg2rad(lon2d);
let u = sin((lat2r - lat1r)/2);
let v = sin((lon2r - lon1r)/2);
return 2.0 * earthRadiusKm * asin(sqrt(u * u + cos(lat1r) * cos(lat2r) * v * v));
}
//test here.... https://andrew.hedges.name/experiments/haversine/
func doTestHaversine(){
let km = distanceEarth(lat1d: 38.898556, lon1d: -77.037852, lat2d: 38.897147, lon2d: -77.043934)
print(km) // should show : 0.549 or similar..
}
回答by Peter Jonsson
This is the method that I would use for finding the distance
这是我用来寻找距离的方法
Or this, not concidering the "bend" of the Earth
或者这个，不考虑地球的“弯曲”
